Properties of the Determinant I

Algebra Level 3

The determinant has the properties:

det ( A B ) = det ( A ) det ( B ) det ( A ) = det ( A T ) \det(AB) = \det(A)\det(B) \\ \det(A) = \det(A^T)

Hence, given that a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 = 16 \begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \\ \end{vmatrix} = 16

Find 4 a 1 2 a 2 4 a 3 2 a 4 a 5 2 a 6 4 a 7 2 a 8 4 a 9 \begin{vmatrix} 4a_1 & 2a_2 & 4a_3 \\ 2a_4 & a_5 & 2a_6 \\ 4a_7 & 2a_8 & 4a_9 \\ \end{vmatrix}

256 256 64 64 4 4 16 16 1 1

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2 solutions

Adam K
Apr 22, 2016

Let us call the first matrix A A and the second B B .

We see we can get B B from A A by applying the following operations:

  1. R 1 2 R 1 R_1 \rightarrow 2R_1
  2. R 3 2 R 3 R_3 \rightarrow 2R_3
  3. C 1 2 C 1 C_1 \rightarrow 2C_1
  4. C 3 2 C 3 C_3 \rightarrow 2C_3

Where R n k R n R_n \rightarrow kR_n multiplies every element in row n n by k k , and Where C n k C n C_n \rightarrow kC_n multiplies every element in column n n by k k . These operations can be represented by the matrices:

E 1 = ( 2 0 0 0 1 0 0 0 1 ) E_1 = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} and E 2 = ( 1 0 0 0 1 0 0 0 2 ) E_2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} .

Where pre-multipying is a row operation and post-multiplying is a column operation.

Then B = E 1 E 2 A E 1 E 2 B = E_1E_2AE_1E_2 , so det ( B ) = det ( E 1 ) 2 det ( E 2 ) 2 det ( A ) = 2 2 2 2 16 = 256 \det(B) = \det(E_1)^2\det(E_2)^2\det(A) = 2^2 \cdot 2^2 \cdot 16 = 256 .

Otto Bretscher
Apr 23, 2016

If we use permutations ("Sarrus' Rule") to find the determinant, we see that every term is multiplied by 16, 16 a 1 a 5 a 9 + . . . 16 a 3 a 5 a 7 16a_1a_5a_9+...-16a_3a_5a_7 , so that the determinant we seek is 1 6 2 = 256 16^2=\boxed{256} .

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